Question

(1 point) 2.0537
An automobile manufacturer would like to know what proportion of its customers are not satisfied with the service provided by the local dealer. The customer relations department will survey a random sample of customers and compute a 96% confidence interval for the proportion who are not satisfied.

(a) Past studies suggest that this proportion will be about 0.22. Find the sample size needed if the margin of the error of the confidence interval is to be about 0.025.
(You will need a critical value accurate to at least 4 decimal places.)
Sample size:

(b) Using the sample size above, when the sample is actually contacted, 15% of the sample say they are not satisfied. What is the margin of the error of the confidence interval?
MoE:

Answer 1

Solution:

Given,

E = 0.025

c = 96% = 0.96

or p = 0.22

1 - 1- p = 1 - 0.22 = 0.78

Now,

= 1 - c = 1 - 0.96 = 0.04

/2 = 0.02

= 2.0540 (using z table)

The sample size for estimating the proportion is given by

n =

= (2.054)2 * 0.22 * 0.78 / (0.0252)

= 1158.3455

=  1159 ..(round to the next whole number)

Answer : Required Sample size is n = 1159

Solution:

Given,

n = 1159 ....... Sample size

   = 0.15

Our aim is to construct 96% confidence interval.

c = 0.96

= 1 - c = 1- 0.96 = 0.04

  /2 = 0.02 and 1- /2 = 0.98

Search the probability 0.98 in the Z table and see corresponding z value

= 2.054   

Now , the margin of error is given by

E =  /2 *  

= 2.054 * [ 0.15 *(1 - 0.15)/ 1159 ]

= 0.0215

The margin of error is 0.0215